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Yes, this is my question...

How can you prove this? That $f(x)=e^x$ from the set of reals to the set of reals is not invertible, but if the codomain is restricted to the set of positive real numbers, the resulting function is invertible. As far as I know, this function is one to one over its entire domain...which means it is invertible.

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Invertible means one-to-one and onto. In particular, we only say that a map $f:A \to B$ is invertible if there is another map $g:B \to A$ such that both $f \circ g$ and $g \circ f$ are the identity maps over their respective spaces.

Of course, any one-to-one map can be made invertible by restricting the codomain to the image of the map. Similarly (assuming the axiom of choice), we can make any onto map invertible by restricting the domain to an appropriate subset.

In some contexts, it makes sense to call the (natural) logarithm the inverse of the exponential map, even when this restriction of the domain is not explicitly stated. As you might expect, the domain of the logarithm must be the positive numbers.

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the function is not surjective as no $-ve$ number has a preimage hence the conclusion

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